Verified Solution • 2025–26 Session moderate

QJEEM23Chf968: JEE Main 2023 Chemistry - Class 12 JEEM23Chf968 Solution: Given below are two statements : Statement I : Ethene at 333 to …

Given below are two statements :

Statement I : Ethene at 333 to 343 K343 \mathrm{~K} and 66 - 77 atm pressure in the presence of AlEt3\mathrm{AlEt}_{3} and TiCl4\mathrm{TiCl}_{4} undergoes addition polymerization to give LDP.

Statement II : Caprolactam at 533533 - 543 K543 \mathrm{~K} in H2O\mathrm{H}_{2} \mathrm{O} through step growth polymerizes to give Nylon 6.

In the light of the above statements, choose the correct answer from the options given below:

Multiple Choice Options

  • A Both Statement I and Statement II are true
  • B Statement I is true but Statement II is false
  • C Statement I is false but Statement II is true
  • D Both Statement I and Statement II are false

Step-by-Step Verified Solution

Statement I is false. Ethene (ethylene) under certain conditions (high pressure, high temperature) in the presence of catalysts does undergo addition polymerization, but it forms polyethylene, not LDP (Low-Density Polyethylene) specifically. The formation of low-density polyethylene (LDPE) vs high-density polyethylene (HDPE) depends on the specific conditions and catalysts used. The catalysts mentioned (AlEt3 and TiCl4) are associated with the production of HDPE through the Ziegler-Natta process, not LDPE.

Statement II is true. Caprolactam polymerizes to form Nylon 6 via a ring-opening polymerization mechanism, which is a type of step-growth polymerization. The process is often carried out at high temperatures, and water is a byproduct of the reaction, so the description in Statement II is accurate.

Therefore, Option C (Statement I is false but Statement II is true) is the correct answer.

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